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Iodine is prepared both in the laboratory and commercially by adding Cl2(g) to an aqueous solution containing sodium iodide. 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) How many grams of sodium iodide, NaI, must be used to produce 87.9 g of iodine, I2?

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Answer:

[tex]m_{NaI}=104gNaI[/tex]

Explanation:

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In this case, given the balanced chemical reaction, it is possible to evidence that the mole ratio of sodium iodide to iodine is 2:1 and the molar masses are 149.89 and 253.81 g/mol respectively; in such a way, we write the following stoichiometric setup in order to obtain the required grams of sodium iodide:

[tex]m_{NaI}=87.9gI_2*\frac{1molI_2}{253.81gI_2}*\frac{2molNaI}{1molI_2}*\frac{149.89gNaI}{1molNaI} \\\\m_{NaI}=104gNaI[/tex]

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